We always say that an object is black to a person's eyes if it absorbs all incident light or doesn't reflect anything, and white if it doesn't absorb and reflects all.
My question is which specific property allows black objects to absorb and stops white objects from absorbing.
Not sure I understand your question. Keeping my answer somewhat colloquial, less technical.
Visible light consists of a mixture of electromagnetic radiation of different wavelengths (=different colors of light). See: Visible Spectrum
Something appears black to us, if no (or only very little) light is coming from it towards our eyes.
Conversely something appears white to us, if a mix of differently colored lights are coming from it towards our eyes.
Some bodies only reflect (or emit) light of one color (or predominantly of one color). These bodies appear in color to us.
A black object is black because it has many internal degrees of freedom that are in the energy range of the visible spectrum. This leads to light in the visible spectrum being absorbed and the energy used to fill those degrees of freedom. In the visible range, those transitions usually correspond to electronic excitations.
First, the perception of colors is biological in our eyes, we have receptors for Red, Green, and Blue wavelength colors. The visible light we see, are combination of these wavelength photons. Our brain perceives them as certain colors, based on the ratio of each (RGB) color in the combination.
Now the easier is black. For our eyes, to see any visible color, we must receive photons to our receptors that are in the visible wavelength. If none of the photons that reach our receptors in our eyes are visible color wavelength, then we will see black. A black object will usually absorb or inelastically scatter (heat up) almost all visible wavelength photons, and will reflect (elastic scattering) almost none of the visible wavelength photons.
Now white is not that easy. White objects are white, because the light that reaches our receptors will use all three (RGB) receptors equally. White light is physically the combination of all visible wavelengths.
When a photon interacts with an atom, three things can happen:
elastic scattering, the photon will keep its energy, and changes angle
inelastic scattering, the photon will give part of its energy to the atom and changes angle
absorption, the photon gives all its energy to the atom and the absorbing electron moves to a higher energy level as per QM
Now for a white object, it is the ratio of these three that cause the object to be white:
most of the photons will be elastically scattered, that is reflection
less photons will be inelastically scattered, that will give thermal energy to the vibrational energy of the atoms and molecules of the metal, it will heat up
absorption, this is very little number of photons will be absorbed, and most of the photons that are absorbed will be re-emitted in multiple steps, so that the re-emitted light will be a combination of all visible wavelengths
Now a white object is white because:
almost all visible wavelength photons will be reflected, and since the sunlight is white (it is not yellow), the white object will just reflect the white light of the Sun, and will seem white. If you shine red light on a white wall, it will seem red, and it works with other colors too
almost all non-visible wavelength photons will be inelastically scattered, and will heat up the metal
very little number of visible wavelength photons are absorbed, and almost all of them will be re-emitted in multiple steps so that the re-emitted light will be a combination of all visible wavelengths, seeming white
So you are right that white objects reflect (elastic scattering) most sunlight (white), and absorb little number of visible wavelength photons. And black objects absorb or inelastically scatter (heat up) almost all visible color wavelength photons, and reflect almost none.
